Sure, let's solve these systems of equations step by step using both the substitution method and the elimination method.
### System 1: [tex]\(x + 2y + 1 = 0\)[/tex] and [tex]\(2x - 3y - 12 = 0\)[/tex]
#### Substitution Method:
1. Solve the first equation for [tex]\(x\)[/tex]:
[tex]\[
x + 2y + 1 = 0 \implies x = -2y - 1
\][/tex]
2. Substitute [tex]\(x = -2y - 1\)[/tex] into the second equation:
[tex]\[
2(-2y - 1) - 3y - 12 = 0
\][/tex]
3. Simplify and solve for [tex]\(y\)[/tex]:
[tex]\[
-4y - 2 - 3y - 12 = 0 \implies -7y - 14 = 0 \implies y = -2
\][/tex]
4. Substitute [tex]\(y = -2\)[/tex] back into [tex]\(x = -2y - 1\)[/tex]:
[tex]\[
x = -2(-2) - 1 = 4 - 1 = 3
\][/tex]
So, the solution to the first system is:
[tex]\[
(x, y) = (3, -2)
\][/tex]
### System 2: [tex]\(2x - y - 3 = 0\)[/tex] and [tex]\(4x + y - 3 = 0\)[/tex]
#### Elimination Method:
1. Add the two equations together to eliminate [tex]\(y\)[/tex]:
[tex]\[
(2x - y - 3) + (4x + y - 3) = 0 \implies 6x - 6 = 0 \implies 6x = 6 \implies x = 1
\][/tex]
2. Substitute [tex]\(x = 1\)[/tex] into the first equation to solve for [tex]\(y\)[/tex]:
[tex]\[
2(1) - y - 3 = 0 \implies 2 - y - 3 = 0 \implies -y - 1 = 0 \implies y = -1
\][/tex]
So, the solution to the second system is:
[tex]\[
(x, y) = (1, -1)
\][/tex]
### Summary of Solutions:
- First system: [tex]\((x, y) = (3, -2)\)[/tex]
- Second system: [tex]\((x, y) = (1, -1)\)[/tex]
These are the solutions for the given systems of equations using the substitution and elimination methods.
